Ch.9 Liquids and Solids

Transcription

1 Ch.9 Liquids and Solids 9.1. Liquid-Vapor Equilibrium 1. Vapor Pressure. Vapor Pressure versus Temperature 3. Boiling Temperature. Critical Temperature and Pressure 9.. Phase Diagram 1. Sublimation. Melting Point 9.3. Molecular Substances; Intermolecular Forces 1. Dispersion (London) Forces. Dipole Forces 3. Hydrogen Forces 9.. Network Covalent; Ionic and Metallic Solids 1. Network Covalent Solids. Ionic Solids 3. Metals 9.5. Crystal Structures 1. Simple Cubic Cell (SC). Face-centered Cubic Cell (FCC) 3. Body-centered Cubic Cell (BCC) 1

2 Ch.9 Liquids and solids Liquid and solid 與 gas 之差異處 : 1. Molecules are much closer to one another in liquids and solids.. Intermolecular forces ( 分子分子間作用力 );which are essentially negligible with gases,but play a much important role in liquids and solids. 液體之表面張力 (surface tension), wetting ability, 可加入 soap or detergent ( 界面活性劑 ) 使其 surface tension wetting ability 9-1 Liquid-vapor equilibrium Vaporization 蒸發 :a liquid converts to a gas (vapor). Open system:liquid 持續蒸發成 vapor 至全部 liquid 用盡 Closed system:liquid 蒸發至一定數量後 equilibrium. liquid vapor dynamic Vapor pressure The pressure exerted by the vapor over the liquid remain constant. 液體上之壓力保持一定. Vapor pressure: 一特定液體在一特定 T 之下有一特定之 vapor pressure 值. Ex: 5 H O 之 vapor pressure = mmhg. 5 ether 之 vapor pressure = 537 mmhg. T=C vapor pressure 較高者 b.p 較低. n 容器大小與 vapor pressure 無關. P = RT v n n/v 固定. v Ex 9-1: A cool-mist vaporizer with capacity.00 L is used to add moisture to dry air in a room at 5. The room has dimensions 1 ft by 1 ft by 8 ft. The vapor pressure of water is mm Hg. Take the density of water at 5 to be 1.00 g/ml. a). If the vaporizer runs until it is empty, what is the vapor pressure of

3 Ans: water in the room? b). How much water is required to completely saturate the air at 5? c). A relative humidity of 33% is desirable in heated space on wintry days. What volume of water is left in the vaporizer when the room s relative humidity reaches that level? (Relative humidity = 100 P/P 0, where P is the actual pressure and P 0 is the vapor pressure at saturation.) L (a) V = (1 1 8) ft = L 3 1 ft 1000 ml 1.00 g 1 mol n HO=.00 L = 111 mol 1 L 1 ml 18.0 g nrt P = = = 0.08 atm = 6 mm Hg V It is impossible to have a vapor pressure larger than mm Hg. Thus, the asumption that all the water is vaporized is wrong. The pressure in the room is mm Hg. PV (/760) ( ) (b) n HO= = = 3 mol RT m = = g HO relative humidity PV (7.9/760) ( ) n HO= = = 1 mol RT m HO = =.5 10 g = 0.5 L 0 (c) P = P = mm Hg =7.9 mm Hg The volume of water left in the vaporizer is: = 1.75 L 3

4 Vapor pressure versus T Vapor pressure of water: 5 mm Hg 50 9 mm Hg mm Hg (a) P = a e H vap / RT ln P vs. 1/T 呈 linear Fig 9- b. y = mx + b ln P = H vap RT + ln a Clausis Clapeyron equation: P H vap 1 1 ln = P1 R T T1

5 P H 1 1 ln vap = P1 R T1 T ΔH vap :J/mol R = 8.31 J/mol.K Ex 9-: Benzene has a vapor pressure of 183 mm Hg at 0. Taking its heat of vaporization to be 30.8 kj/mol, calculate its vapor pressure at 5. Ans:P 1 = 183 mm Hg; T 1 = 313 K P =? ; T = 98 K ; H vap = 30.8 kj/mol P H vap 1 1 ln = P 1 R T1 T 3 P ln = = ( ) P ln = P = P = 101 mm Hg Boiling point (b.p.): T b A liquid boils at a temperature at which its vapor pressure is equal to the pressure above its surface. T b (b.p) α P Normal boiling point: 正常沸點 vapor pressure of a liquid = 760 mm Hg water 之 normal boiling point = 100 at mmhg 水之 b.p. = 5 Critical temperature and pressure: CO (l) 0 vapor pressure 3 atm 10 atm 0 56 atm atm T>31 the tube contains only one phase vapor 5

6 When T>31,no matter how much pressure is applied,it is impossible to have liquid CO. 即使 P = 1000atm, 35 or 0 之溫度均無法使 CO 液化. Critical temperature: 臨界溫度 The temperature above which the liquid phase of a pure substance cannot exist. The pressure that must be applied to cause condensation at that temperature is called the critical pressure. Critical pressure is the vapor pressure of the liquid at the critical temperature. CO 之 T c = 31 P c = 73 atm T c < 5 6

7 Supercritical fluid 超臨界流體 : 超過臨界溫度及壓力之物質 have unusual solvent properties. 例 :Supercritical CO 萃取 coffee 中之 caffeine. 9- Phase diagram: P vs. T Fig 9-5:Phase diagram of H O AB:liquid-vapor equilibrium line AC:vapor-solid equilibrium line AD:liquid-solid equilibrium line A: triple point 三相點 ; for H O 0.01,.56 mm Hg (l)(s)(g) 三相共存 Ex 9-3: Consider a sample of H O at point A in Figure 9-5. a). what phase(s) is (are) present? b). If the temperature of the sample were reduced at constant pressure, what would happen? c). How would you convert the sample to vapor without changing the temperature? Ans: a. A 為 triple point 同時呈現 gas liquid 及 solid 三相. 7

8 b. 成為 solid ( ice). c. T =C 使成為 vapor P 下降, P<.56 mm Hg Sublimation 昇華 Solid changes directly to vapor without passing through the liquid phase. 一固體要發生昇華 T<T 三相點才可若 T>T 三相點固體會融化成為 liquid. 昇華 : 先將 T<T 三相點, 再逐步降壓至 Fig. 9.5 中 s-g 平衡線 (AC) 以下即可發生昇華. 碘 奈丸 Melting point 熔點 s l Identical to freezing point ( 凝固點 ) l s. The temperature at which solid and liquid phase are in equilibrium. for most substance,the melting point at 1atm is virtually identical with the triple point temp. H O: melting point 0.00 triple point 0.01 差

9 P 對 m.p. 之影響 : 1. 如 Fig 9-7(a): (s)-(l) 之平衡線向右傾 固態更具凝相 佔絕大多數 ; P m.p. 一般化合物多屬此種. 如 Fig 9-7(b): (s)-(l) 之平衡線向左傾 液態更具凝相 佔少數 ; P m.p. 如 : 水. 壓力增加 13 atm, melting point 下降低 1. 9